Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{k + 7}{-8k + 32} \div \dfrac{k^2 + 10k + 21}{k - 4} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{k + 7}{-8k + 32} \times \dfrac{k - 4}{k^2 + 10k + 21} $ First factor the quadratic. $z = \dfrac{k + 7}{-8k + 32} \times \dfrac{k - 4}{(k + 7)(k + 3)} $ Then factor out any other terms. $z = \dfrac{k + 7}{-8(k - 4)} \times \dfrac{k - 4}{(k + 7)(k + 3)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (k + 7) \times (k - 4) } { -8(k - 4) \times (k + 7)(k + 3) } $ $z = \dfrac{ (k + 7)(k - 4)}{ -8(k - 4)(k + 7)(k + 3)} $ Notice that $(k - 4)$ and $(k + 7)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ \cancel{(k + 7)}(k - 4)}{ -8(k - 4)\cancel{(k + 7)}(k + 3)} $ We are dividing by $k + 7$ , so $k + 7 \neq 0$ Therefore, $k \neq -7$ $z = \dfrac{ \cancel{(k + 7)}\cancel{(k - 4)}}{ -8\cancel{(k - 4)}\cancel{(k + 7)}(k + 3)} $ We are dividing by $k - 4$ , so $k - 4 \neq 0$ Therefore, $k \neq 4$ $z = \dfrac{1}{-8(k + 3)} $ $z = \dfrac{-1}{8(k + 3)} ; \space k \neq -7 ; \space k \neq 4 $